If 25 mL of 2.5 M HCl are titrated with 2.2 M NaOH, what is the pH after adding 15 mL NaOH

Equation:
NaOH+ HCl → NaCl + H2O
1mol NaOH reacts with 1 mol HCl

Mol HCl in 25mL of 2.5M HCl solution = 25/1000*2.5 = 0.0625 mol HCl
Mol NaOH in 15mL of 2.2M NaOH = 15/1000*2.2 = 0.033 mol NaOH
When reacted there is 0.0625-0.033 = 0.0295 mol HCl remaining unreacted, dissolved in 40mL solution.
Molarity of HCl solution = 0.0295/0.040 = 0.7375M HCl

because HCl is a strong acid:
[H+] = [HCl] = 0.7375M

pH = - log [H+]
pH = - log 0.7375
pH = 0.13

pH of the final solution = 0.13

First you have to know the HCl is a strong acid ... dissociates completely
HCl --> H+ + Cl- ( That means that the number of moles of H+ = HCl )


number of moles of HCl = c*v = 0.025*2.5 = 0.0625mol = n(H+)
number of moles of NaOH added = c*v=2.2*0.015=0.033mol = n(OH-) ADDED
the number of moles pf OH- added are gonna react with the moles of H+
final number of moles of H= 0.0625-0.033=0.0295mol
The final volume = 40ml=0.040L
Concentration of H+ = n/v = 0.0295/0.040 = 0.7375M
pH=-log[0.7375]= 0.1322

25ml of 2.5M HCl = .0625 moles H+

15ml of 2.2M NaOH = .033 moles OH-

OH- and H+ cancel out on a 1:1 ratio, so just subtract moles OH- from moles H+

.0625-.033 = .0295 moles H+

This means you have .0295 moles H+ per 40ml of solution (add the 15ml and 25ml)

pH = -log[H+] so .0295/.04 = [.7375H+]. -log[.7375] = pH of .13